In the lab, a chemist adds 3.6 grams of sodium chloride to 100.0 g of water. what is the molality of the solution?

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superman1987 superman1987 · Answer 1 · 2019-06-05T15:44:24+02:00

Answer:

0.616 m.

Explanation:

Molality (m) is defined as the no. of moles of solute dissolved in a 1.0 kg of the solvent.

m = (no. of moles of solute)/(mass of the solvent (kg))

m = (mass/molar mass) NaCl / (mass of the solvent (kg))

mass of NaCl = 3.6 g, molar mass of NaCl = 58.44 g/mol, mass of water = 100.0 g = 0.1 kg.

m = (mass/molar mass) NaCl / (mass of the solvent (kg)) = (3.6 g / 58.44 g/mol) / (0.1 kg) = 0.616 m.

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samueladesida43 samueladesida43 · Answer 2 · 2022-03-31T12:58:01+02:00

The molality of a solution if a chemist adds 3.6 grams of sodium chloride to 100.0 g of water is 0.615m.

Molality is defined as the no. of moles of solute dissolved in a 1.0 kg of the solvent.

m = no. of moles of solute ÷ mass of the solvent (kg)

no of moles of NaCl = 3.6g ÷ 58.5g/mol = 0.0615mol

Molality = 0.0615mol ÷ 0.1kg

Molality = 0.615m

Therefore, the molality of a solution if a chemist adds 3.6 grams of sodium chloride to 100.0 g of water is 0.615m.

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